Question

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QUESTION 10 Red blood cell count is distributed normally with a mean of 5.4 million cells /ulL of blood and a standard deviation of 0.7 million cells /uL of blood. What is the probability of a healthy adult having less than 7 million cells /uL of blood?

Answer 1

Given that, mean $(\mu)$ = 5.4 million cells /uL

standard deviation $(\sigma)$ = 0.7 million cells /uL

Here, $X \sim Normal (\mu = 5.4, \sigma=0.7)$

We want to find, P( X < 7 million cells /uL)

$P(X <7) = P(\frac { X-\mu}{\sigma}< \frac { 7-5.4}{0.7}) =P(Z <2.29)=0.9890$

P(X < 7) = 0.9890

Therefore, the probability of a healthy adult having less than 7 million cells /uL of blood is 0.9890

Note: using standard normal z-table we get,

P(Z < 2.29) = 0.9890